Stoichiometry 1 PDF

Stoichiometry 1 Stoichiometry is the branch of chemistry that tells us the relationship between the amounts of reactants and products in a chemical reaction. Chemists need a good knowledge of stoichiometry because it enables them to decide on the quantities of reactants needed to form a certain amou...

Stoichiometry 1 Stoichiometry is the branch of chemistry that tells us the relationship between the amounts of reactants and products in a chemical reaction. Chemists need a good knowledge of stoichiometry because it enables them to decide on the quantities of reactants needed to form a certain amount of product. To Calculate the Percentage of each Element in a Compound Calculate the percentage composition by mass of each element in calcium pxide, CaO. Calculate the relative molecular mass of CaO. Ca = 40 O = 16 Rel. Molecular Mass = 56 40 Percentage of Ca = 56 x 100 = 71.43% 16 Percentage of O = 56 x 100 = 28.57% The percentage composition of CaO by mass is 71.43% Ca and 28.57% O. To Calculate the Empirical Formula of a Compound Given the Percentage Composition **Definitions: The molecular formula of a compound is a formula which shows the number and type of each atom present in a molecule of that compound. The empirical formula of a compound is the formula showing the simplest whole number ratio of the numbers of different atoms present in the molecule.** Molecular formula = empirical formula x n (where n is a whole number) The empirical formula of glucose is CH2O. When the relative molecular mass of glucose is measured by a mass spectrometer, it is found to be 180. What is the molecular formula of glucose? Molecular formula = empirical formula x n Molecular formula = (CH2O) x n 180 = (12 + 2 + 16) x n 180 = 30 x n n=6 (CH2O) x 6 = C6H12O6 1 To calculate the Empirical Formula of a Compound Given the Masses of Reactants and Products In an experiment to determine the formula of an oxide of copper, hydrogen gas was passed over the heated oxide. It was found that 5.4 g of the oxide gave 4.8 g of copper. Calculate the empirical formula of the oxide. Copper oxide + hydrogen → copper 5.4 g 4.8 g Mass of copper oxide = 5.4 g Mass of copper in copper oxide = 4.8 g Therefore, mass of oxygen in copper oxide = 5.4 – 4.8 = 0.6 g 4.8 No. moles of copper atoms in copper oxide = 63.5 = 0.076 moles 0.6 No. moles of oxygen atoms in copper oxide = = 0.038 moles 16 Ratio of Cu:O = 0.076:0.038 = 2:1 Empirical formula of copper oxide is Cu2O Thus, the following reaction has occurred Cu2O + H2 → 2Cu + H2O Calculating Masses of Reactants or Products from Balanced Chemical Equations Magnesium reacts with oxygen to produce magnesium oxide according to the equation 2Mg + O2 → 2MgO If a student burns 9 g of magnesium in excess oxygen, what mass of magnesium oxide will be formed? 2Mg + O2 → 2 MgO 2Mg → 2MgO 2 moles → 2 moles 2(24) g → 2 (24 + 16) g 48 g Mg 80g MgO 80 Therefore, 1 g Mg = 48 g MgO 9 𝑥 80 9g Mg g MgO 58 15 g of magnesium oxide will be formed from 9 g of Mg. 2 Calculating Volumes of Gases from Balanced Chemical Equations Carbon dioxide is formed when carbon is burned in air according to the equation. C + O → CO2 What volume of carbon dioxide is formed when 30 g of carbon are burned in excess oxygen? C → CO2 1 mole → 1 mole 12 g C → 22.4 L of CO2 at s.t.p 22.4 1gC → L of CO2 12 30 𝑥 22.4 30 g C → L of CO2 12 Answer: 56L of Carbon dioxide 3